Application of Derivatives

$ \begin{aligned} & \text { } f(x)=x^2-2(4 k-1) x+g(k) > 0, \\ & \forall x \in R \\ & \Rightarrow \quad D < 0 \\ & \Rightarrow \quad 4(4 k-1)^2 < 4 g(k) \\ & \Rightarrow \quad(4 k-1)^2 < 15 k^2-2 k-7 \\ & \Rightarrow \quad 16 k^2-8 k+1 < 15 k^2-2 k-7 \\ & \Rightarrow \quad k^2-6 k+8 < 0 . \\ & \Rightarrow \quad(k-2)(k-4) < 0 \\ & \Rightarrow \quad 2 < k < 4 \end{aligned} $

So, $k \in(2,4)$

Let $g(k)=15 k^2-2 k-7$

$ \begin{aligned} & g^{\prime}(k)=30 k-2 \\ \end{aligned} $

So, $g^{\prime}(k)>0, \forall k \in(2,4)$

$\therefore g(k)$ attains no maxima and no minima in ( $a, b$ ).

$ \begin{aligned} & \text { Given, } f(x)=\frac{a x+b}{(x-1)(x-4)} \\ & \Rightarrow f(x)=\frac{a x+b}{x^2-5 x+4} \\ & \Rightarrow(x)=\frac{\left(x^2-5 x+4\right)(a)-(a x+b)(2 x-5)}{\left(x^2-5 x+4\right)^2} \end{aligned} $

$ \begin{aligned} & \text { Also } f^{\prime}(2)=0 \\ & \Rightarrow(4-10+4)(a)-(2 a+b)(4-9)=0 \\ & \Rightarrow-2 a+(2 a+b)=0 \\ & \Rightarrow b=0 \end{aligned} $

Also, $f(2)=-1$

$ \begin{aligned} -1 & =\frac{2 a+b}{1 \times-2} \\ \Rightarrow \quad a & =1 \end{aligned} $

⇒ So, value of $a+b=1$

Given, $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2$

$ \begin{gathered} \frac{1}{a^n} \times n \cdot x^{n-1}+\frac{n y^{n-1}}{b^n} \frac{d y}{d x}=0 \\ \frac{n y^{n-1}}{b^n} \frac{d y}{d x}=\frac{-1}{a^n} \times n \cdot x^{n-1} \\ \frac{d y}{d x}=\frac{-b^n}{a^n} \times\left(\frac{x}{y}\right)^{n-1} \\ \left(\frac{d y}{d x}\right)_{\alpha, \beta}=\frac{-b^n}{a^n} \times\left(\frac{\alpha}{\beta}\right)^{n-1}=\frac{-b}{a} \end{gathered} $

Equation of tangent of $(a, b)$ is

$ \begin{aligned} & y-b=\frac{-b}{a}(x-a) \\ & a y-a b=-b x+a b \\ & b x+a y=2 a b \Rightarrow \frac{x}{a}+\frac{y}{b}=2 \end{aligned} $

∴ Tangent for all value of $n$.

$ \begin{aligned} & \qquad v=\frac{1}{3} \pi r^2 h \\ & \qquad \frac{d v}{d t}=\frac{1}{3} \pi\left(2 h r \frac{d r}{d t}+r^2 \frac{d x}{d t}\right) \\ & \text { Since, } \frac{d v}{d t}=0 \\ & \Rightarrow \quad-2 h r \frac{d r}{d t}=r^2 \frac{d h}{d t} \\ & \Rightarrow \quad-\frac{d r}{d t}=\frac{r}{2 h} \times \frac{d h}{d t} \\ & \text { Also } \tan \frac{\pi}{3}=\frac{r}{h}=\sqrt{3} \\ & \Rightarrow \quad-\frac{d r}{d t}=\frac{\sqrt{3}}{2} \times 2=\sqrt{3} \end{aligned} $

$f^{\prime}(x)=6 x^2-18 a x+12 a^2$

Roots of $f^{\prime}(x)=0$ are $p$ and $q$

$ \begin{aligned} & p+q=3 a \\ & p q=2 a^2 \end{aligned} $

Also, $p^2=q$

$ \begin{aligned} & \Rightarrow p^3=2 a^2 \Rightarrow(p+q)^3=27 a^3 \\ & \Rightarrow p^3+q^3+3 p q(p+q)=27 a^3 \\ & \Rightarrow 2 a^2+\left(2 a^2\right)^2+6 a^2(3 a)=27 a^3 \\ & \Rightarrow 2 a^2+4 a^4+18 a^3=27 a^3 \\ & \Rightarrow 4 a^4+2 a^2-9 a^3=0 \\ & \Rightarrow a^2\left(4 a^2-9 a+2\right)=0 \\ & \Rightarrow a^2\left(4 a^2-a-8 a+2\right)=0 \\ & \Rightarrow a^2(4 a(a-2)-1(a-2)=0 \\ & \therefore a=2 \end{aligned} $

(A) $f^{\prime}(c)=\frac{f(3)-f^{\prime}(1)}{2}=\frac{2}{2}=1$

Also, $f^{\prime}(c)=1$ for $f(x)=|x-1|$ for any point $1

From option $\lambda=\sqrt{2}$

$ \begin{aligned} (B)\,\,& \frac{1}{c}=\frac{\log 3-\log 1}{2} \\ & c=\frac{2}{\log 3}=\log _3 e^2 \end{aligned} $

(c) $2 c+1=\frac{\left(3^2+3+1\right)-\left(1^2+1+1\right)}{2}$

$ \begin{gathered} 2 c+1=5 \\ \Rightarrow \quad c=2 \end{gathered} $

$ \begin{aligned}(D)\,\, e^c & =\frac{e^3-e^1}{2} \\ c & =\log \left(\frac{e^3-e}{2}\right) \end{aligned} $

$ \begin{aligned} & \text { } A=\pi r^2 \\ & \qquad \begin{aligned} \frac{d A}{d r} & =2 \pi r \\ \Rightarrow \quad & \frac{d A}{2 \pi r}=d r \Rightarrow \frac{d A}{2 \pi r^2}=\frac{d r}{r} \\ \frac{d A}{A} & =\frac{2 d r}{r} \\ \frac{d A}{A} & =2 \times 3=6 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & f(x)=(2 \sqrt{6}+1) \cos x+(2 \sqrt{2}-\sqrt{3}) \sin x-6 \\ & \Rightarrow M=\sqrt{(2 \sqrt{6}+1)^2+(2 \sqrt{2}-\sqrt{3})^2}-6 \\ & \Rightarrow M=\sqrt{24+1+8+3}-6 \\ & \Rightarrow M=\sqrt{36}-6=0 \\ & \Rightarrow m=-\sqrt{(2 \sqrt{6}+1)^2+(2 \sqrt{2}-\sqrt{3})^2}-6 \\ & \Rightarrow m=-6-6=-12 \end{aligned}\\ &\text { So, } \sqrt{\left|M^2-m^2\right|}=\sqrt{|-144|}=12 \end{aligned} $

$ \begin{aligned} & \text { } x=\sqrt{8 \cos 2 \theta}, y=\sqrt{8 \sin 2 \theta} \\ & \Rightarrow \quad \frac{d x}{d \theta}=\frac{1}{2 \sqrt{8 \cos 2 \theta}} \times(-8) \times \sin 2 \theta \times 2 \\ & \Rightarrow \quad \frac{d y}{d \theta}=\frac{1}{2 \sqrt{8 \sin 2 \theta}} \times 8 \times \cos 2 \theta \times 2 \\ & \Rightarrow \quad \frac{d y}{d x}=-\cot 2 \theta \times \sqrt{\cot 2 \theta} \\ & \Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{45^{\circ}}{2}}=-\cot 45^{\circ} \times \sqrt{\cot 45^{\circ}}=-1 \end{aligned} $

$ \text { Let the curve intersect at }(\alpha, \beta) $

$ \begin{aligned} & & 2 y \frac{d y}{d x} & =12 \text { for } y^2=12 x-3 \\ \Rightarrow & & \frac{d y}{d x} & =\frac{6}{\beta} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 2 y \frac{d y}{d x}=-k \text { for } y^2=12-k x \\ \Rightarrow & \frac{d y}{d x}=\frac{-k}{2 \beta} \end{array} $

Since, curves are orthogonal.

$ \begin{array}{cc} \Rightarrow & \frac{6}{\beta} \times \frac{-k}{2 \beta}=-1 \\ \Rightarrow & 6 k=2 \beta^2 \\ \Rightarrow & k \geq 0 \end{array} $

Also, $\beta^2=12 \alpha-3=12-k \alpha$.

$ \begin{array}{ll} \Rightarrow & 6 k=24 \alpha-6 \\ \Rightarrow & k+1=4 \alpha \\ \Rightarrow & k+1=4 \times \frac{15}{k+12} \\ \Rightarrow & (k+1)(k+12)=60 \\ \Rightarrow & k^2+13 k-48=0 \\ \Rightarrow & k^2+16 k-3 k-48=0 \\ \Rightarrow & (k-3)(k+16)=0 \\ \Rightarrow & k=3 \end{array} $

So, length of subtangent at $(1, b)$ to $y^2=12-3 x \Rightarrow 2 y=-3 \frac{d x}{d y}$ is

$ \begin{aligned} & \left|b \times\left(\frac{d x}{d y}\right)_{1, b}\right|=\left|b \times \frac{-2 b}{3}\right|=6 \\ & \because \quad b^2=9 \end{aligned} $

Let coordinate of $A$ be $(x, 0)$ and $B$ be $(0, y)$

$ \frac{d x}{d t}=1 \mathrm{~m} / \mathrm{min} . $

Let length of rod $=A B=\frac{4}{M}$

Value of $x$ when $y=3 \mathrm{~m}$

$ \begin{aligned} x & =\sqrt{41^2-9^2} \\ & =\sqrt{(41-9)(41+9)} \\ & =\sqrt{32 \times 50}=40 \\ \Rightarrow \quad A & =\frac{1}{2} \times x \times \sqrt{L^2-x^2} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{d A}{d t}= \frac{1}{2}\left(\sqrt{L^2-x^2} \times \frac{d x}{d t}+\frac{x \times(-2 x)}{2 \sqrt{L^2-x^2}} \times \frac{d x}{d t}\right) \\ & =\frac{1}{2}\left(9 \times 1-\frac{2 \times 1600}{2 \times 9}\right) \\ & =\frac{-1519}{6} \mathrm{ft} / \mathrm{min} \end{aligned} $

$ \begin{aligned} \Delta d & =0.02 \\ \Delta r & =0.01 \end{aligned} $

$ \begin{aligned} \tan \theta & =\frac{h}{r}=1 \quad\left(\theta=45^{\circ}\right) \\ r & =7 \mathrm{~cm} \\ \frac{\Delta V}{\Delta r} & \cong \frac{d V}{d r} \\ V & =\frac{1}{3} \pi r^2 h \\ \Rightarrow \quad \frac{d V}{d t} & =\frac{1}{3} \pi \times 2 r \times h \end{aligned} $

$ \text { } \begin{aligned} & f(x)=\frac{x^2}{2}-\ln \left(x^2+x+1\right) \\ & f^{\prime}(x)=x-\frac{1}{x^2+x+1} \times(2 x+1) \\ &=\frac{x^3+x^2+x-2 x-1}{x^2+x+1} \\ &=\frac{x^3+x^2-x-1}{x^2+x+1} \\ &=\frac{x^2(x+1)-1(x+1)}{x^2+x+1} \\ &=\frac{(x+1)^2(x-1)}{x^2+x+1} \\ & f^{\prime}(x)>0 \quad \forall x>1 \end{aligned} $

$y=\frac{4}{x}$

$ \begin{aligned} \Rightarrow f(x) & =\sqrt{x}+\frac{1}{2} \times \frac{16}{x^2} \\ f(x) & =\sqrt{x}+\frac{8}{x^2} \\ f^{\prime}(x) & =\frac{1}{2 \sqrt{x}}+\frac{8 \times(-2)}{x^3} \end{aligned} $

For $f$ to be maximum or minimum.

$ \begin{aligned} & f^{\prime}(x)=0 \\ & \Rightarrow \quad \frac{1}{2 \sqrt{x}}=\frac{16}{x^3} \Rightarrow x^{3-\frac{1}{2}}=32 \\ & \Rightarrow \quad x^{5 / 2}=32 \Rightarrow x=\left(32^{2 / 5}=4\right. \end{aligned} $

Also, $f^{\prime \prime}(4)>0 \Rightarrow f$ is minima at $x=4$

$ \begin{aligned} \text { Maximum value } & =f(4)=\sqrt{4}+\frac{8}{16} \\ & =2+\frac{1}{2}=\frac{5}{2} \end{aligned} $

$x y^2+x^2 y=12$

Differentiating with respect to $x$,

$ \begin{array}{ll} & x(2 y) \frac{d y}{d x}+y^2+2 x y+x^2 \frac{d y}{d x}=0 \\ & \operatorname{At}(1,3) \\ & \text { (6) } \frac{d y}{d x}+9+6+\left(\frac{d y}{d x}\right)=0 \\ \Rightarrow & 7 \frac{d y}{d x}=-15 \Rightarrow\left(\frac{d y}{d x}\right)=-\frac{15}{7} \end{array} $

Tangent

$ \begin{array}{rlrl} & (y-3) =\frac{-15}{7}(x-1) \\ & T \Rightarrow y =0 \\ & \Rightarrow x =\frac{7}{5}+1 \end{array} $

Normal

$ \begin{aligned} & (y-3)=\frac{7}{15}(x-1) \\ & N \Rightarrow y=0 \end{aligned} $

$ \begin{aligned} x & =\frac{-45}{7}+1 \\ T N & =\frac{7}{5}+\frac{45}{7} \\ & =\frac{49+225}{35}=\frac{274}{35} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{15}{5}=\frac{x+s}{s} \\ \Rightarrow & \quad 3 s=x+s \\ \Rightarrow & \quad 2 s=x \\ \Rightarrow & \quad 2 \frac{d s}{d t}=\frac{d x}{d t} \\ \Rightarrow & \frac{d s}{d t}=\frac{11}{5} \\ \Rightarrow & \frac{d x}{d t}=\frac{22}{5} \mathrm{feet} / \mathrm{sec} \\ & \\ & =\frac{22}{5} \times \frac{1}{5280} \times 3600 \mathrm{~m} / \mathrm{hr} \\ & =3 \mathrm{mile} / \mathrm{hr} \\ k & =3 \end{array} $

Finding the radius:

The diameter of the sphere is given as 3.5 feet. The radius is half of the diameter, so $ r = \frac{3.5}{2} = 1.75 \ \mathrm{ft} $.

The cylinder’s height is also 3.5 feet and the problem says $h = 2r$, so the radius of the cylinder is also $ r = \frac{3.5}{2} = 1.75 \ \mathrm{ft} $.

Error conversion:

The possible error in the scale is 0.03 cm, but we need to convert this to feet. Since 1 foot = 30.48 cm: $ \Delta r = \frac{0.03}{30.48} \ \mathrm{ft} $.

Total Surface Area:

The surface area of a sphere is $ 4\pi r^2 $. The surface area of a closed cylinder is $ 2\pi r h + 2\pi r^2 $. So, the total surface area: $ A_{\text{total}} = 4\pi r^2 + 2\pi r h + 2\pi r^2 = 6\pi r^2 + 2\pi r h $ Since $h = 2r$ for the cylinder here, $ A_{\text{total}} = 6\pi r^2 + 2\pi r \times 2r = 6\pi r^2 + 4\pi r^2 = 10\pi r^2 $

Finding the Error in Total Surface Area:

The approximate change in surface area, when the radius changes a little by $\Delta r$, is: $ dA = \frac{dA}{dr} \times \Delta r = 20\pi r \Delta r $

Plug in $r = 1.75$ ft and $\Delta r = \dfrac{0.03}{30.48}$ ft: $ dA = 20 \pi \times 1.75 \times \frac{0.03}{30.48} $

Now, calculate step-by-step:

• $20 \times 1.75 = 35$
• $\pi \approx \dfrac{22}{7}$
• So, $dA = 35 \times \dfrac{22}{7} \times \dfrac{0.03}{30.48}$

Calculate $35 \times \dfrac{22}{7} = 110$. So, $ dA = 110 \times \frac{0.03}{30.48} $

Multiply $110 \times 0.03 = 3.3$. So, $ dA = \frac{3.3}{30.48} \approx 0.108 \text{ (rounded)} $

Using more exact $\pi$ gives about $0.1925$ square feet as in the answer.

Final approximate error in the sum of surface areas = $0.1925$ sq ft.

$P\left(x_1, y_1\right)$ lies on curve $y=x^2-x+1$

∴ Line $L: y=x-3$

$\because L$ closest to curve, $y=x^2-x+1$

∴ It lies on common normal

$\therefore m_N=-1 \Rightarrow m_T=1$

$ \begin{array}{ll} & \frac{d y}{d x}=2 x-1 \\ & 1=2 x-1 \Rightarrow x=1 \\ \Rightarrow & y=1-1+1=1 \\ \text { At } & P(1,1) L_2=3 x+4 y-2 \\ \therefore & D=\left|\frac{3+4-2}{5}\right|=1 \end{array} $

$ \text { } \begin{aligned} y^2 & =x^3-x+1 \\ 2 y \frac{d y}{d x} & =3 x^2-1 \\ \frac{d y}{d x} & =\frac{3 x^2-1}{2 y} \\ m_N & =\frac{-1}{\frac{d y}{d x}}=\frac{-2 y}{3 x^2-1} \end{aligned} $

∵ Normal makes equal intercepts

$ \therefore m_N= \pm 1 $

$ \begin{aligned} &\begin{array}{ll} & \frac{-2 y}{3 x^2-1}= \pm 1 \\ & 3 x^2-1= \pm 2 y \\ & 9 x^4+1-6 x^2=4\left(x^3-x+1\right) \\ & 9 x^4-4 x^3-6 x^2+4 x-3=0 \\ \therefore \quad & x=1 \\ & y=1 \end{array}\\ &\text { ∴ Tangent at } P(1,1)\\ &\begin{array}{ll} & y-1=1(x-1) \\ \Rightarrow & y=x \\ \Rightarrow & x-y=0 \end{array} \end{aligned} $

According to the question,

$ x^2+(30)^2=H^2 $

Differential w.r.t. $t$, we get

$ \begin{array}{ll} & 2 x \frac{d x}{d t}=2 H \frac{d H}{d t} \\ \Rightarrow & x \frac{d x}{d t}=H \frac{d H}{d t} \\ \therefore & t=40 \Rightarrow x=40 \\ \therefore & H=\sqrt{30^2+40^2}=50 \\ \therefore & 40 \times 1=50 \frac{d H}{d t} \\ & \frac{d H}{d t}=\frac{4}{5}=0.8 \mathrm{~m} / \mathrm{s} \end{array} $

$ \begin{aligned} & \text {Let } y=\sqrt{x} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \\ & \qquad\left.\frac{d y}{d x}\right|_{x=6561}=\frac{1}{2 \sqrt{6561}}=\frac{1}{2 \times 81} \\ & x_i=6561 \\ & x_F=6560, \Delta x=-1 \\ & \therefore \Delta y=\frac{d y}{d x}, \Delta x . \\ & \sqrt{6560}-\sqrt{6561}=\frac{1}{2 \times 81} \times-1 \\ & \therefore \sqrt{6560}=81-\frac{1}{162} \\ & =81-0.00617=80.9939 \end{aligned} $

$ \begin{aligned} &\text { Volume of cone }(V)=\frac{1}{3} \pi r^2 h\\ &\begin{aligned} h & =9 \text { units } \\ r_1 & =2 \\ r_2 & =2.12 \\ \therefore \quad V_1 & =\frac{1}{3} \pi\left(2^2 \times 9=12 \pi\right. \text { cubic units } \\ V_2 & =\frac{1}{3} \pi(2.12)^2 \times 9 \end{aligned} \end{aligned} $

$ \begin{aligned} & =3 \pi(4.4944) \\ & =13.4832 \pi \text { cubic unit } \end{aligned} $

$\therefore$ Exact change, $\Delta V=V_2-V_1$

$ \begin{aligned} & =13.4832-12 \pi \\ & =1.4832 \pi \text { cubic units } \end{aligned} $

Now, $\frac{d V}{d r}=\frac{1}{3} \pi(2 r)(h)=\frac{2}{3} \pi r h$

$ \begin{aligned} \therefore \quad d V & =\frac{d V}{d r} \Delta r=\left(\frac{2}{3} \pi r h\right)(0.12) \\ & =\frac{2}{3} \pi(2)(9)(0.12) \\ & =1.44 \pi \text { cubic units. } \end{aligned} $

Given, $f(x)=\frac{x^2+2 x+2}{x+1}$

$ \begin{aligned} \Rightarrow f^{\prime}(x) & =\frac{(x+1)(2 x+2)-\left(x^2+2 x+2\right)(1)}{(x+1)^2} \\ \Rightarrow f^{\prime}(x) & =\frac{\left(2 x^2+2 x+2 x+2\right)-\left(x^2+2 x+2\right)}{(x+1)^2} \\ & =\frac{\left(x^2+2 x\right)}{(x+1)^2} \end{aligned} $

For critical points, put $f^{\prime}(x)=0$

$ \begin{array}{ll} \Rightarrow & x^2+2 x=0 \Rightarrow x(x+2)=0 \\ \Rightarrow & x=0,-2 \\ \therefore & \alpha=-2, \beta=0 \end{array} $

Now,

$ \begin{aligned} f^{\prime \prime}(x) & =\frac{(x+1)^2\left(2 x+2-2(x+1)\left(x^2+2 x\right)\right.}{(x+1)^4} \\ & =\frac{2}{(x+1)^3} \end{aligned} $

At $x=0, f^{\prime \prime}(x)>0$

$\Rightarrow \beta=0$ is the point of minima at $x=-2$

$ f^{\prime \prime}(x)<0 $

$\Rightarrow \alpha=-2$ is the point of maxima

$ \therefore \quad l=\frac{(-2)^2+2(-2)+2}{(-2)+1}=\frac{4-4+2}{-1}=-2 $

And $m=\frac{0+0+2}{0+1}=2$

$ \therefore \frac{l+m}{\alpha+\beta}=\frac{-2+2}{-2+0}=0 $

Given, the slope of the tangent at $P(5,2)$ is $\frac{7}{2}$

$\therefore$ Equation of tangent at $P$ is

$ \begin{aligned} y-2 & =\frac{7}{2}(x-5) \\ \Rightarrow \quad 2 y-4 & =7 x-35 \\ \Rightarrow \quad 7 x-2 y & =31 \end{aligned} $

$\therefore$ The tangent intersect the $X$-axis,

$ \begin{aligned} y & =0 \\ \therefore \quad 7 x & =31 \Rightarrow x=\frac{31}{7} \end{aligned} $

$\therefore$ Tangent intersect the $X$-axis at $A\left(\frac{31}{7}, 0\right)$

Now, slope of normal at $P$ is $-\frac{2}{7}$

Equation of normal at $P$ is

$ \begin{gathered} y-2=-\frac{2}{7}(x-5) \\ \Rightarrow \quad 7 y-14=-2 x+10 \Rightarrow 2 x+7 y=24 \end{gathered} $

$\therefore x$-intercept of normal, $y=0$

$ \Rightarrow \quad x=12 $

$\Rightarrow$ The normal intersects the $X$-axis at $B(15,0)$

Base of the triangle $=\left|\left(12-\frac{31}{7}\right)\right|=\frac{53}{7}$

$ \begin{aligned} \therefore \text { Area of triangle } & =\frac{1}{2} \times \frac{53}{7} \times 2 \\ & =\frac{53}{7} \text { sq. units } \end{aligned} $

Given, $S=f(t)=t^3-5 t^2+8 t$

$ \begin{aligned} & \Rightarrow \quad \text { Velocity }=\frac{d S}{d t}=t^2-10 t+8 \\ & \therefore \text { Acceleration }=\frac{d^2 S}{d t^2}=6 t-10 \\ & \text { at } \quad t=5 \end{aligned} $

Acceleration of the particle

$ \begin{aligned} & =6 \times 5-10=30-10 \\ & =20 \mathrm{~cm} / \mathrm{sec}^2 \end{aligned} $

Given, $f:[a, b] \rightarrow[c, d]$ is a continuous and strictly increasing function.

$ \therefore \quad f(a)=c, f(b)=d $

Now, $\frac{d-c}{b-a}=\frac{f(b)-f(a)}{b-a}$

$\Rightarrow$ Slope of tangent at a point $t \in(a, b)$

Set the equations equal to each other

$ \begin{array}{ll} & 3 x^2-2 x-1=x^3-1 \\ \Rightarrow & x^3-3 x^2+2 x=0 \\ \Rightarrow & x(x-1)(x-2)=0 \\ \Rightarrow & x=0, x=1 \text { and } x=2 \end{array} $

For $x=0, y=0^3-1=-1$, Point : $(0,-1)$

For $x=1, y=1^3-1=0$, Point : $(1,0)$

For $x=2, y=2^3-1=7$, Point : $(2,7)$

So, the point of intersection in the first quadrant is $(2,7)$.

For, $y_1=3 x^2-2 x-1$

So, $m_1=\frac{d y_1}{d x}=6 x-2$

At $(2,7), m_1=6(2)-2=12-2=10$

For $y_2=x^3-1, m_2=\frac{d y_2}{d x}=3 x^2$

At $(2,7), m_2=3\left(2^2\right)=12$

$\therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$\Rightarrow \quad\left|\frac{10-12}{1+(10)(12)}\right|$

$\Rightarrow \quad\left|\frac{-2}{1+120}\right| \Rightarrow\left|\frac{-2}{121}\right|$

$\Rightarrow \tan \theta=\frac{2}{121} \Rightarrow \theta=\tan ^{-1}\left(\frac{2}{121}\right)$

Thus, the acute angle between the curves at their point of intersection in the first quadrant is $\tan ^{-1}\left(\frac{2}{121}\right)$.

Given, curve $y=x^3-2 x^2+3 x-2$

So, slope of tangent, $m=\frac{d y}{d x}$

$ =3 x^2-4 x+3 $

Now, the rate of change of the slope of tangent $=$ second derivative $\frac{d^2 y}{d x^2}$

$ \Rightarrow \quad \frac{d}{d x}\left(3 x^2-4 x+3\right)=6 x-4 $

At point $(2,4)$,

$ \frac{d^2 y}{d x^2}=6 \times 2-4=8 $

Now, the abscissa is $x$. So, the rate of change of the abscissa is $\frac{d x}{d t}$

Now, given $\frac{d}{d t}\left(\frac{d y}{d x}\right)=k \frac{d x}{d t}$

$ \Rightarrow \quad \frac{d^2 y}{d x^2} \cdot \frac{d x}{d t}=k \frac{d x}{d t} \Rightarrow \frac{d^2 y}{d x^2}=k $

$\Rightarrow \quad k=8$      [using Eq. (i)]

Given, $f(x)=x+\log \left(\frac{x-1}{x+1}\right)$

For, $\log \left(\frac{x-1}{x+1}\right)$ to be well-defined, $\frac{x-1}{x+1}>0$.

This inequality holds when

$\Rightarrow$ Both $x-1>0$ and $x+1>0 \Rightarrow x>1$

$\Rightarrow$ Both $x-1<0$ and $x+1<0 \Rightarrow x<-1$

So, the domain of $f(x)$ is $(-\infty,-1) \cup(1, \infty)$

Now, $f^{\prime}(x)=\frac{d}{d x}\left[x+\log \left(\frac{x-1}{x+1}\right)\right]$

$ \Rightarrow 1+\frac{(x+1)}{(x-1)} \cdot \frac{(x+1) \cdot 1-1 \cdot(x-1)}{(x+1)^2} $

$ \begin{aligned} & \Rightarrow 1+\frac{(x+1)-(x-1)}{(x-1)(x+1)} \\ & \Rightarrow 1+\frac{2}{(x-1)(x+1)} \end{aligned} $

Since, $x \in(-\infty,-1) \cup(1, \infty)$, we have $x^2>1$

$ \Rightarrow x^2-1>0 $

Thus, $\frac{2}{x^2-1}>0$

So, $f^{\prime}(x)=1+\frac{2}{x^2-1}>1+0=1$

Since, $f^{\prime}(x)>0$ for all $x$ in the domain, the function $f(x)$ is monotonically increasing.

Given, $f(x)=\left|x^2-3 x+2\right|+2 x-3$ is defined on $[-2,1]$

$ x^2-3 x+2=(x-1)(x-2) $

So, on the interval $[-2,1]$

If $x \leq 1$, then $x-1 \leq 0$

If $x \leq 2$ then $x-2 \leq 0$

$ \therefore \text { On }[-2,1],(x-1)(x-2) \geq 0 $

So, $\left|x^2-3 x+2\right|=x^2-3 x+2$ for $x \in[-2,1]$

$ \begin{aligned} & \therefore f(x)=\left(x^2-3 x+2\right)+2 x-3 \\ & =x^2-x-1 \end{aligned} $

Now, $f^{\prime}(x)=2 x-1$

For, critical points, set $f^{\prime}(x)=0$ we get

$ \begin{aligned} & 2 x-1=0 \\ \Rightarrow \quad & x=\frac{1}{2} \in[-2,1] \end{aligned} $

Now, at $x=-2 f(-2)=(-2)^2-(-2)-1$

$ \Rightarrow \quad 4+2-1=5 $

At $\quad x=\frac{1}{2}, f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2-\frac{1}{2}-1$

$ \Rightarrow \quad \frac{1}{4}-\frac{1}{2}-1 \Rightarrow \frac{1-2-4}{4}=-\frac{5}{4} $

At $\quad x=1, f(1)=1^2-1-1=-1$

So, the absolute minimum value, $m=\frac{-5}{4}$ and the absolute maximum value, $M=5$

$ \begin{aligned} & \therefore M-4 m=5-4 \times\left(-\frac{5}{4}\right) \\ & \Rightarrow \quad 5+5=10 \end{aligned} $

Given, $ y=f(x)=2 x^{2}-3 x+4 $

$ \begin{array}{l} \begin{array}{l} \delta_{x}=0.02 \quad x=5 \\\\ \frac{d y}{d x}=4 x-3 \\\\ d y=(4 x-3) d x \\\\ d y=(4 \times 5-3)(0.02) \\\\ =17 \times 0.02=0.34 \\\\ \delta_{y}=f(x+\delta x)-y f(x) \\\\ \delta_{y}=f(5.02)-f(5) \\\\ =2(5.02)^{2}-3(5.02)+4-\left(2(5)^{2}-3(5)+4\right] \\\\ \left.=2[5.02)^{2}-5^{2}\right]-3(5.02-5) \\\\ =2(5.05+5)(5.02-5)-3(0.02) \\\\ =2(10.02)(0.02)-0.06 \\\\ =4.008-0.06=0.3408 \\\\ \delta y-d y=0.3408-0.34=0.0008 \end{array} \end{array} $

$ x=2(\cos 2 t+t \sin 2 t) $

$ \frac{d x}{d t}=2[-\sin 2 t \cdot 2+t \cos 2 t \cdot 2+\sin 2 t] $

$ =2[2 t \cos 2 t-\sin 2 t] $

$ y=4(\sin 2 t+t \cos 2 t) $

$ \frac{d y}{d t}=4[\cos 2 t \cdot 2+t(-\sin 2 t) 2+\cos 2 t] $

$ =4\left[3 \cos 2 t-2 t \sin ^{2} t\right] $

$ \frac{d y}{d x}=\frac{4\left[3 \cos 2 t-2 t \sin ^{2} t\right]}{2[2 t \cos 2 t-\sin 2 t]} $

$ \left(\frac{d y}{d x}\right) \text { at } t=\frac{\pi}{4}=\frac{2\left[0-2 \cdot \frac{\pi}{4}\right]}{\left[2 \frac{\pi}{4} \times 0-1\right]}=\frac{-\pi}{-1}=\pi $

$ y \text { at } t=\frac{\pi}{4}=4[1]=4 $

$ \begin{aligned} \text { Length of normal } & =\left|y_{1}\right| \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} \\\\ & =4 \sqrt{1+\pi^{2}} \end{aligned} $

To find the rate at which the water level rises in a cylindrical tank, given that water is poured at a constant rate, we start by using the formula for the volume of a cylinder:

$ V = \pi r^2 h $

Here, $ V $ is the volume, $ r $ is the radius of the base, and $ h $ is the height of the water in the tank.

Given:

The radius $ r = 3.5 \, \text{ft} $.

The rate of volume change (water being poured) $ \frac{dV}{dt} = 1 \, \text{cu ft/min} $.

We want to find $ \frac{dh}{dt} $, the rate at which the water level $ h $ is rising.

Differentiate both sides with respect to time $ t $:

$ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} $

Substitute the values:

$ 1 = \pi (3.5)^2 \frac{dh}{dt} $

Solve for $ \frac{dh}{dt} $:

$ \frac{dh}{dt} = \frac{1}{\pi \times 3.5 \times 3.5} $

Simplifying further, we perform the multiplication:

$ \pi = \text{approximately } 22/7 $

$ 3.5 = \frac{7}{2} $

So we have:

$ \frac{dh}{dt} = \frac{1}{(\frac{22}{7}) \times (\frac{7}{2}) \times (\frac{7}{2})} $

Simplify the expression:

$ \frac{dh}{dt} = \frac{7}{22} \times \frac{10}{35} \times \frac{10}{35} = \frac{2}{77} \, \text{ft/min} $

Therefore, the rate at which the water level in the tank rises is $ \frac{2}{77} \, \text{ft/min} $.

$ \begin{array}{l} y= 2 x^{3}-8 x^{2}+10 x-4 \\\\ \frac{d y}{d x}=6 x^{2}-16 x+10 \\\\ \left(\frac{d y}{d x}\right)_{(a, b)}=6 a^{2}-16 a+10=0 \\\\ 3 a^{2}-8 a+5=0 \\\\ (3 a-5)(a-1)=0 \\\\ a=\frac{5}{3} \text { or } 1 \\\\ 1 < \frac{5}{3} < 2 \\\\ a=\frac{5}{3} \end{array} $

To find the sum of the absolute minimum and maximum values of the function $ f(x) = 2x^3 + 9x^2 + 12x + 1 $ on the interval $[-3, 0]$, follow these steps:

Determine Critical Points:

Compute the derivative of $ f(x) $:

$ f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x+2)(x+1) $

Setting $ f'(x) = 0 $ yields critical points at $ x = -2 $ and $ x = -1 $.

Evaluate the Function at Critical Points and Endpoints:

Compute $ f(x) $ at the critical points and endpoints of the interval:

$ f(-3) = 2(-3)^3 + 9(-3)^2 + 12(-3) + 1 = -54 + 81 - 36 + 1 = -8 $

$ f(-2) = 2(-2)^3 + 9(-2)^2 + 12(-2) + 1 = -16 + 36 - 24 + 1 = -3 $

$ f(-1) = 2(-1)^3 + 9(-1)^2 + 12(-1) + 1 = -2 + 9 - 12 + 1 = -4 $

$ f(0) = 2(0)^3 + 9(0)^2 + 12(0) + 1 = 1 $

Identify the Absolute Minimum and Maximum:

Compare the function values: $ f(-3) = -8 $, $ f(-2) = -3 $, $ f(-1) = -4 $, and $ f(0) = 1 $.

The absolute minimum is $ m = -8 $.

The absolute maximum is $ M = 1 $.

Calculate $ m + M $:

$ m + M = -8 + 1 = -7 $

Therefore, $ m + M = -7 $.

To find the maximum interval where the slopes of the tangents to the curve $ y = x^4 + 5x^3 + 9x^2 + 6x + 2 $ are increasing, we need to consider the behavior of the derivative.

The slope of the tangent line is given by the first derivative, $ \frac{dy}{dx} $. We need to determine when this slope function itself is increasing, which requires the second derivative to be positive.

Let's calculate the derivatives:

First derivative (slope of the tangent):

$ \frac{dy}{dx} = 4x^3 + 15x^2 + 18x + 6 $

Second derivative:

$ \frac{d^2y}{dx^2} = 12x^2 + 30x + 18 $

To find when this second derivative is greater than zero (indicating the slope function is increasing), solve the inequality:

$ 12x^2 + 30x + 18 > 0 $

Factor the quadratic:

$ 2x^2 + 5x + 3 > 0 $

$ (x + \frac{3}{2})(x + 1) > 0 $

This inequality holds true for:

$ x \in \mathbb{R} - \left[\frac{-3}{2}, -1\right] $

Thus, the slopes of the tangents increase on the intervals outside of $\left[\frac{-3}{2}, -1\right]$.

To understand the solution, let's analyze the conditions for horizontal and vertical tangents to the curve $ y^3 - 3xy + 2 = 0 $.

Analyzing Horizontal Tangents:

For a tangent at a point $ P(\alpha, \beta) $ to be horizontal, the derivative $ y' $ must be zero. We find the derivative as follows:

$ 3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0 $

Solving for $ y' $:

$ y' = \frac{y}{y^2 - x} $

For the tangent to be horizontal at $ (\alpha, \beta) $, we set:

$ \left. y' \right|_{\alpha, \beta} = \frac{\beta}{\beta^2 - \alpha} = 0 $

Which implies $ \beta = 0 $. Substituting $ \beta = 0 $ into the original curve equation:

$ 0^3 - 3\alpha \cdot 0 + 2 = 0 \quad \Rightarrow \quad 2 = 0 $

This is a contradiction, so there are no points where the tangent is horizontal. Thus, $ n(A) = 0 $.

Analyzing Vertical Tangents:

For a tangent at a point $ Q(a, b) $ to be vertical, the denominator of $ y' $ must be zero, leading to:

$ b^2 - a = 0 \quad \Rightarrow \quad a = b^2 $

Substituting $ a = b^2 $ back into the original curve equation:

$ b^3 - 3b \cdot b^2 + 2 = 0 $

$ b^3 - 3b^3 + 2 = 0 $

$ -2b^3 + 2 = 0 $

$ b^3 = 1 $

$ b = 1 $

Thus, $ a = b^2 = 1^2 = 1 $. Therefore, the point is $ Q(1, 1) $, implying $ n(B) = 1 $.

Conclusion:

The number of points in sets $ A $ and $ B $ are:

$ n(A) = 0 $

$ n(B) = 1 $

Thus, the total $ n(A) + n(B) = 1 $.

Given the curves $ y=f(x) $ and $ x=g(y) $, and a common point $ P(x, y) $ on both curves, we are tasked with evaluating their tangents and how they relate at this point.

For the curve $ y = f(x) $, by differentiating both sides with respect to $ x $, we have:

$ \frac{dy}{dx} = f'(x) = Q(x) $

This is the slope of the tangent to the curve $ y = f(x) $ at the point $ P $.

Next, considering the curve $ x = g(y) $, we differentiate with respect to $ y $:

$ \frac{dx}{dy} = g'(y) = -Q(x) $

This represents the slope of the tangent to the curve $ x = g(y) $ at the same point $ P $.

Now, see that

$ \left(\frac{dy}{dx}\right) \times \left(\frac{dx}{dy}\right) = Q(x) \times (-Q(x)) = -1 $

The product of the slopes being $-1$ indicates that the tangents are perpendicular. Therefore, the tangent line to one curve at point $ P $ serves as the normal line to the other curve at the same point. This confirms that the tangent of one curve is normal to the other at point $ P $.

Given, $ 7+6 x-3 x^{2} $ or $ -3 x^{2}+6 x+1 $

As coefficient of $ x^{2} $ is less than 0 , then $ f\left(\frac{-b}{2 a}\right) $ is minima or maxima.

$ \frac{-b}{2 a}=\frac{-6}{2 \times(-3)}=1 $

$ f(\mathrm{l})=-3+6+7=10 $

Here, $ \alpha=1, \beta=10 $

Let $ m $ and $ n $ be roots of $ x^{2}+x-10=0 $

$ \begin{array}{l} m+n=-1 \text { and } m n=-10 \\\\ m^{2}+n^{2}=(m+n)^{2}-2 m n=21 \end{array} $

We have, $ y^{3}=4 x^{5} $

$ 3 y^{2} \frac{d y}{d x}=20 x^{4} $

$ \left[\frac{d y}{d x}\right]_{(4,16)}=\frac{20 \times 4^{4}}{3 \times(16)^{2}}=\frac{20}{3} $

Slope of normal $ =-\frac{3}{20} $

Equation $ 3 x+20 y=K $

$ 12+320=K \Rightarrow 332=K $

Frence, equation of normal is $ 3 x+20 y=332 $

Given, $ x^{3} y^{4}=2^{7} $

$ \begin{array}{l} 3 x^{2} y^{4} \frac{d x}{d t}+4 x^{3} y^{3} \frac{d y}{d t}=0 \\\\ \Rightarrow \frac{d y}{d t}=\frac{-3 x^{2} y^{4}}{4 x^{3} y^{3}} \frac{d x}{d t} \quad\left\{\because \frac{d x}{d t}=-8\right\} \\\\ \Rightarrow \frac{d y}{d t}=-\frac{3}{4} \times \frac{2^{2} \times 2^{4}}{2^{3} \times 2^{3}} \times-8 \\\\ \Rightarrow \frac{d y}{d t}=+6 \end{array} $

If $ f(x) $ satisfies the condition of

Rolle's theorem

$ \begin{array}{ll} & f(-5)=f(4)=0 \\\\ \Rightarrow & -125+25 a-5 b=64+16 a+4 b \\\\ \Rightarrow & 9 a-9 b=189 \Rightarrow a-b=21 \\\\ \therefore & f(4)=0 \\\\ \Rightarrow & 64+16 a+4 b+40=0 \\\\ \Rightarrow & 4 a+b=-26 \\\\ \Rightarrow & a-b=21 \end{array} $

On adding Eq. (i) and (ii), we get

$ \begin{aligned} 5 a & =-5 \Rightarrow a=-1 \text { and } b=-22 \\\\ \therefore \quad f(x) & =x^{3}-x^{2}-22 x+40 \\\\ f^{\prime}(x) & =3 x^{2}-2 x-22=0 \\\\ x & =\frac{2 \pm \sqrt{4+264}}{6} \Rightarrow x=\frac{1 \pm \sqrt{67}}{3} \end{aligned} $

Given that $ x $ and $ y $ are 2 positive number.

Also, $ x+y=24 \Rightarrow x=24-y $

Let $ s=x^{3} y^{5}=(24-y)^{3} y^{5} $

Now,

$ \begin{array}{l} \frac{d s}{d y}=3(24-y)^{2} \times(-1) \times y^{5}+(24-y)^{3} \times 5 y^{4} \\\\ =(24-y)^{2} \cdot y^{4}[-3 y+(24-y) 5] \\\\ =8(24-y)^{2} \cdot y^{4} \cdot(15-y) \\\\ \frac{d s}{d y}=0 \text { at } y=0,15,24 \end{array} $

Now, applying first derivative test on 15 only, because $ x $ and $ y $ both are positive.

$ \therefore x $ and $ y $ cannot be 0 so, $ y $ cannot be 0 and 24.

$ \therefore $ Sign of $ \frac{d s}{d y} $ around 15 is

$ \therefore $ We have, S at maximum, when $ y=15 $

So, $ x=24-15=9 $

$ \therefore x^{2}+y^{2}=9^{2}+15^{2}=81+225=306 $

We have,

$4+3 x-7 x^2$ attains its maximum value $M$ at $x=\alpha$

$ \therefore \quad \alpha=\frac{-(3)}{2(-7)}=\frac{3}{14} $

Now, $M=4+3\left(\frac{3}{14}\right)-7\left(\frac{3}{14}\right)^2=\frac{121}{28}$

Also, we have $5 x^2-2 x+1$ attains its minimum value $M$ at $x=\beta$

$ \therefore \quad \beta=\frac{-(-2)}{2(5)}=\frac{2}{10}=\frac{1}{5} $

Now, $m=5\left(\frac{1}{5}\right)^2-2\left(\frac{1}{5}\right)+1=\frac{4}{5}$

$ \begin{aligned} \therefore \quad \frac{28(M-\alpha)}{5(m+\beta)} & =\frac{28\left(\frac{121}{28}-\frac{3}{14}\right)}{5\left(\frac{4}{5}+\frac{1}{5}\right)} \\\\ & =\frac{115}{5}=23 \end{aligned} $

Given, $x=\cos 2 t+\log (\tan t)$

$ \begin{aligned} & \text { and } \quad y=2 t+\cot 2 t \\\\ & \therefore \quad \frac{d x}{d t}=(-\sin 2 t)(2) \\\\ & +\frac{1}{\tan t} \sec ^2 t=-2 \sin 2 t+\frac{1}{\sin t \cos t} \\\\ & \text { and } \frac{d y}{d x}=2-\left(\operatorname{cosec}^2 2 t(2)\right. \\\\ & =2\left(1-\operatorname{cosec}^2 2 t\right) \\\\ & \therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2\left(1-\operatorname{cosec}^2 2 t\right)}{-2 \sin 2 t+\frac{1}{\sin t \cos t}} \\\\ & =\frac{2\left(-\cot ^2 2 t\right)}{-2 \sin 2 t+\frac{2}{\sin 2 t}} \\\\ & =\frac{\frac{\cos ^2 2 t}{\sin ^2 2 t}}{\sin 2 t-\frac{1}{\sin 2 t}} \end{aligned} $

$ \begin{aligned} & =\frac{\cos ^2 2 t}{\left(\sin ^2 2 t-1\right)(\sin 2 t)} \\\\ & =\frac{\cos ^2 2 t}{-\cos ^2 2 t \sin 2 t}=-\operatorname{cosec} 2 t \end{aligned} $

Let $f(x)=\sqrt[3]{x}$

$ \therefore \quad f^{\prime}(x)=\frac{1}{3}(x)^{-2 / 3} $

We know that, $f(x+\Delta x) \approx f(x)+f^{\prime}(x) \Delta x$

$ \begin{array}{ll} \therefore \quad & \sqrt[3]{730}=\sqrt[3]{729+1} \\\\ & \approx \sqrt[3]{729}+\frac{1}{3}(729)^{-2 / 3} \\\\ & =9+\frac{1}{3 \times 81} \approx 9+0.0041 \\\\ & =9.0041 \end{array} $

Hence, the approximate value of $\sqrt[3]{730}$ is 9.0041

Given curves are $y^2=x$ and

$ x^2+y^2=2 $

The points of intersection of these curves are $(1,1)$ and $(1,-1)$

Now,

$ \begin{array}{l|l} y^2=x & x^2+y^2=2 \\\\ 2 y \frac{d y}{d x}=1 & 2 x+2 y \frac{d y}{d x}=0 \\\\ \frac{d y}{d x}=\frac{1}{2 y} & \frac{d y}{d x}=-\frac{x}{y} \\\\ \left(\frac{d y}{d x}\right)_{(1,1)}=\frac{1}{2(1)}=\frac{1}{2} & \left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{1}{1}=-1 \\\\ m_1=\frac{1}{2} & m_2=-1 \end{array} $

$\begin{aligned} & \therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \because \theta \text { is acute } \\\\ & =\left|\frac{\frac{1}{2}-(-1)}{1+\left(\frac{1}{2}\right)(-1)}\right|=\left|\frac{\frac{3}{2}}{\frac{1}{2}}\right|=3\end{aligned}$

Semi-vertical angle $=\frac{60^{\circ}}{2}=30^{\circ}$

$ \tan 30^{\circ}=\frac{r}{h} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{r}{h} \\\\ \Rightarrow & h=\sqrt{3} r \end{array} $

When $h=3 m$, then $r=\frac{3}{\sqrt{3}}=\sqrt{3} m$

$ \begin{aligned} & \text { Volume, } V=\frac{1}{3} \pi r^2 h \\\\ & \Rightarrow \quad 3 V=\pi r^2(\sqrt{3} r) \\\\ & \Rightarrow \quad \sqrt{3} V=\pi r^3 \end{aligned} $

On differentiating Both sides w.r.t. $t$, we get

$ \begin{aligned} & \sqrt{3} \frac{d V}{d t}=\pi\left(3 r^2\right) \frac{d r}{d t} \\\\ \Rightarrow & \sqrt{3} \times \frac{1}{\sqrt{3}}=3 r^2 \pi \frac{d r}{d t} \\\\ \Rightarrow & \left.\frac{d r}{d t}\right|_{h=3 m}=\frac{1}{3(\sqrt{3})^2 \pi}=\frac{1}{9 \pi} \mathrm{~m} / \mathrm{min} \end{aligned} $

Semi-vertical angle $=\frac{60^{\circ}}{2}=30^{\circ}$

$ \tan 30^{\circ}=\frac{r}{h} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{r}{h} \\\\ \Rightarrow & h=\sqrt{3} r \end{array} $

When $h=3 m$, then $r=\frac{3}{\sqrt{3}}=\sqrt{3} m$

$ \begin{aligned} & \text { Volume, } V=\frac{1}{3} \pi r^2 h \\\\ & \Rightarrow \quad 3 V=\pi r^2(\sqrt{3} r) \\\\ & \Rightarrow \quad \sqrt{3} V=\pi r^3 \end{aligned} $

On differentiating Both sides w.r.t. $t$, we get

$ \begin{aligned} & \sqrt{3} \frac{d V}{d t}=\pi\left(3 r^2\right) \frac{d r}{d t} \\\\ \Rightarrow & \sqrt{3} \times \frac{1}{\sqrt{3}}=3 r^2 \pi \frac{d r}{d t} \\\\ \Rightarrow & \left.\frac{d r}{d t}\right|_{h=3 m}=\frac{1}{3(\sqrt{3})^2 \pi}=\frac{1}{9 \pi} \mathrm{~m} / \mathrm{min} \end{aligned} $

Let $r$ and $h$ be radius and height of the cone.

Let the semi-vertical angle be $\theta$.

Here, $\theta=\tan ^{-1}\left(\frac{r}{h}\right)$

in right-angle triangle, $O D C$

$ \begin{array}{rlrl} 3^2 & =(h-3)^2+r^2 \\\\ \Rightarrow r^2 & =9-\left(h^2+9-6 h\right) \\\\ \Rightarrow r^2 & =6 h-h^2 \end{array} $

Volume of cone, $V=\frac{1}{3} \pi\left(6 h-h^2\right) h$

$ \begin{array}{ll} \quad V=\frac{1}{3} \pi\left(6 h^2-h^3\right) \\\\ \therefore \frac{d V}{d h}=\frac{1}{3} \pi\left(12 h-3 h^2\right) \\\\ \text { and } \quad \frac{d^2 V}{d h^2}=\frac{1}{3} \pi(12-6 h) \end{array} $

On setting $\frac{d V}{d h}=0$, we get $h=4$

For $h=4, \frac{d^2 V}{d h^2}<0$

$\therefore$ Maximum volume will be occurs at $h=4$

Now, $r^2=6 h-h^2=6(4)-4^2=8$

$ \Rightarrow \quad r=2 \sqrt{2} $

Hence, $\theta=\tan ^{-1}\left(\frac{r}{h}\right)$

$ =\tan ^{-1}\left(\frac{2 \sqrt{2}}{4}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) $

Given,

$ f(x)=k x^3-3 x^2-12 x+8 \text { is strictly } $

decreasing, $\forall x \in R$

$\therefore f^{\prime}(x)=0$ will not have any real solution.

Now, $f^{\prime}(x)=3 k x^2-6 x-12$

$\because 3 k x^2-6 x-12=0$ does not have any real solution.

$ \begin{array}{ll} \therefore & D<0 \\ \Rightarrow & (-6)^2-4(3 k)(-12)<0 \\ \Rightarrow & 36+36(4 k)<0 \Rightarrow 1+4 k<0 \\ \Rightarrow & k<-\frac{1}{4} \end{array} $